Vedanta Optional Mathematics Teacher's Guide ~ 10 201 its slope is, m1 =– Coefficient of x Coefficient of y =– 1 –1 =1 slope of line joining points (4,6) and (10,12), m2 = y2 -y1 x2 -x1 = 12–6 10–4 = 6 6 =1 Here, m1 =m2 , hence the line (i) and the line joining points (4,6) and (10,12) are parallel. 7, Show that lines joining the points (7, –5) and (3,4) is perpendicular to the line 4x–2y +7=0. Solution Slope of line joining points (7,–5) and (3,4) is given by, m= y2 -y1 x2 -x = 1 4+5 3-7 = 9 –4 =– 9 4 Slope of the line 4x –9y+7=7 is m2 =– 4 -9 = 4 9 now, m1 . m2 = 9 -4 . 4 9 =–1 Since the product of slopes is –1, the lines are perpendicular to each other. 8. Find the value of k is that the lines represented by kx +3y+5 =0 and 4x=3y+10=0 are perpendicular to each other. Solution Here given equations of lines are kx +3y+5 =0 ...........(i) and 4x=3y+10=0.............(ii) slope of line (i), m1 =– Coefficient of x Coefficient of y =– k 3 Slope of line, m2 =– 4 -3 = 4 3 As the two lines are perpendicular to each other m1 .m2 =–1 or,– k 3 . 4 3 =–1 or, k= 9 4 9. Find the equation of the straight line that passes through the point (4,5) and perpendicular to the line having slope 6 5 . Solution Let the required line be y – y1 =m(x – x1 ) It passes through the point (4,5) y – 5 = m(x – 4).........(i)
202 Vedanta Optional Mathematics Teacher's Guide ~ 10 slope ot given line is (m1 ) = 6 5 As the lines are perpendicular to each other m.m1 = –1 or, m. 6 5 =–1 m = – 5 6 put the value of m is equation (i), we get y – 5=– 5 6 (x – 4) or, 6y – 30=–5x +20 5x + 6y =50 which is the required equation. 10. (a) Find the equation of straight line which passes through the point (2,3) and parallel to the line x – 2y – 2=0 . Solution Given equations of straight line is x – 2y – 2=0 its slope is (m1 ) = – Coefficient of x Coefficient of y =– 1 -2 = 1 2 we find the equation of straight line which is parallel to given line. So its slope (m)=m1 = 1 2 Now, equation of required line passing through (2,3) and with slope m is given by y – y1 =m(x–x1 ) or, y – 3 = 1 2 (x–2) or, 2y – 6 =x–2 or, x – 2y+4=0 x–2y+4=0 is the required equation. (b) Find the equation if the straight line which passes through the point of intersection of the straight line 3x+4y=7 and 5x–2y=3 and perpendicular to the line 2x+3y=5. Solution For the point of intersection of the straight lines 3x+4y=7 and 5x–2y=3, from 1st equation, and from 2nd equation 3x+4y=7 5x–2y=3 or, x = 7-4y 3 or, x= 3+2y 5 from both 7-4y 3 = 3+2y 5 or, 35–20y=9 + 6y
Vedanta Optional Mathematics Teacher's Guide ~ 10 203 y = 1 now, x=3+2×1 5 = 1 Let (x,y)=(1,1) be (x1 ,y1 ) and the slope of required line be m, m ×slope of equation (2x+3y=5)=–1 (pepr to each other) or, m× -2 3 =–1 m= 3 2 now, eqn. of st. lines is, y–y1 =m(x–x1 ) or, y – 1= 3 2 (x –1) or, 2y – 2=3x–3 or, 3x–2y–1 =0 3x–2y–1=0 is the required equation. (c) Find the equation if the straight line which passes through the point of intersection of 2x – 3y +1=0 and x+2y=3 and parallel to the line 4x+3y=12. Solution For the point of intersection of the straight lines 2x – 3y +1=0 and x+2y=3 from 1st equation, and from 2nd equation 2x – 3y =–1 x=3–2y or, x = 3y-1 2 from both 3y-1 2 = 3–2y or, 3y–1=6 – 4y y = 1 Also, x=3–2×1=1 Let (x,y)=(1,1) be (x1 ,y1 ) for slope (m), two lines passing through (1,1) and 4x+3y=12 are parallel. Now, eqn. of straight lines is, y–y1 =m(x–x1 ) or, y – 1= -4 3 (x –1) or, 3y – 3=–4x+4 4x+3y=7 is the required equation. (d) Find the equation of the straight line that divides the line joining the points p(–1,–4) and
204 Vedanta Optional Mathematics Teacher's Guide ~ 10 Q(7,1) in the ratio of 3:2 and perpendicular to it. Solution The given two points are P(–1,–4) and Q(7,1) Slope of PQ (m1 )= y2 -y1 x2 -x1 = 1+4 7+1 = 5 8 Let R(x,y) be the point of PQ which divides at in 3:2 ratio. (x,y)= m1 x2 +m2 x1 m1 +m2 , m1 y2 +m2 y1 m1 +m2 = 3.7+2.(-1) 3+2 , 3.1+2.(-4) 3+2 = 19 5 ,-1 Since the required line is perpendicular to PQ and passes through 19 5 ,-1 , its slope (m2 ) is given by m1 .m2 =–1 or, 5 8 .m2 =–1 or, m2 =– 8 5 Now, required equation is y–y1 =m(x–x1 ) y +1=– 8 5 x19 5 or, 40x+25y=127 40x+25y=127 is the required equation. (e) Find the equation of the straight line which passes through the centroid of ∆ABC with vertices A(4,5), B(–4,–5) and C(1,2) and parallel to 7x +5y=35. Solution For centroid of ∆ABC, G(x,y)= x1 +x2 +x3 3 , y1 +y2 +y2 3 = 4-4+1 3 , 5-5+2 3 = 1 3 , 2 3 Let m be the slope of required line which is perpendicular to 7x +5y=35 and passes through G 3 R 2 P(–1,–4) Q(7,1) B(–4,–5) C(1,2) 7x+5y=35 A(4,5) G
Vedanta Optional Mathematics Teacher's Guide ~ 10 205 1 3 , 2 3 m =slope of eqn. (7x +5y=35) [ parallel to each other] m2 =– 7 5 Now, required equation is y–y1 =m(x–x1 ) y – 2 3 =– 7 5 x- 1 3 or, 5y– 10 3 =–7x+ 7 3 or, 7x+5y= 17 3 21x+15y=17 is the required equation. 11. If the angle between the lines (a2 –b2 )x–(p+q)y=0 and (p2 –q2 )x+(a+b)y=0 is 90˚, prove that (a–b)(p–q)=1. Solution Given equations of the lines are (a2 –b2 )x–(p+q)y=0 and ...........(i) and (p2 –q2 )x+(a+b)y=0.............(ii) From equation (i),slope(m1 ) = – Coefficient of x Coefficient of y =– (a2 -b2 ) –(p+q) = a2 -b2 p+q From equation (ii), slope (m2 )==– (p2 -q2 ) a+b Since the lines (i) and (ii) are perpendicular to each other m1 .m2 =–1 or, a2 -b2 p+q . p2 -q2 a+b - =–1 or, (a+b)(a-b)(p-q)(p+q) (p+q)(a+b) = 1 (a–b)(p–q) = 1 proved 12.(a) From the point p(–2,4), perpendicular PQ is drawn to the line AB:7x–4y+15=0. Find the equation of PQ. Solution Equation of AB is 7x–4y+15=0 Its Slope is (m1 )= 7 (-4) = 7 4 PQ is perpendicular to AB, A Q B P(–2,4)
206 Vedanta Optional Mathematics Teacher's Guide ~ 10 So slope of PQ (m2 )=– 4 7 ( m1 .m2 =–1) Now the equaton of line PQ is given by y–y1 =m(x–x1 ) y – 4= –4 7 (x+2) 4x+7y=20 is the required equation. 12(b). Find the equation of the altitude PM drawn from the vertex P to QR in ∆PQR p(2,3),Q(–4,1) and R(2,0). Solution Here PM is perpendicular to QR. Slope of line QR (m2 )= y2 –y1 x2 –x1 = 0–1 2+4 = 1 –6 = –1 6 Since they PM is perpendicular to QR perpendicular to QR, m1 ×m2 =–1, where m2 is the slope of PM + 1 6 ×m2 =+1 m2 =6 Equation of PM is y–y1 =m(x–x1 ) y – 3=6(x–2) y – 3=6x–12 or, 6x–y–9=0 6x–y=9 is the required equation. 13. Find the equation of the line perpendicular bisector to the join of the following two points. (a) P(–3,5) and Q(–6,7) Solution Given points are P(–3,5) and Q(–6,7) Mid point PQ=M x1 +x2 2 , y1 +y2 2 =M -3-6 2 , 5+7 2 =M -9 2 , 6 Q(–4,1) R(2,0) P(2,3) M M Q(–6,7) N P(–3,5)
Vedanta Optional Mathematics Teacher's Guide ~ 10 207 Slope of PQ (m1 )= y2 -y1 x2 -x1 = 7-5 -6+3 =– 2 3 Let MN be the perpendicular bisector of PQ with slope(m2 ). then, m1 – m2 = –1 or, – 2 3 .m2 =–1 m2 = 3 2 Now, equation of MN is y–y1 =m(x–x1 ), (⸪ m=m2 =3 2 ) or, y – 6= 3 2 x + 9 2 or, 4y – 24=6x +27 or, 6x–4y+51=0 6x–4y +51=0 is the required equation. (b) P(5,6) and Q(7,10) Solution We have to find the equation of perpendicular bisector of MN Mid point PQ=M x1 +x2 2 , y1 +y2 2 =M 5+7 2 , 6+10 2 =(6,8) Slope of MN is: m1 = y2 –y1 x2 –x1 = 10–6 7–5 = 4 2 =2 Let RS be the perpendicular of MN Since the lines are perpendicular, slope of MN=m2 is given by m1 × m2 = –1 or, – 2 3 .m2 =–1 R(6,8) N(7,10) S M(5,6)
208 Vedanta Optional Mathematics Teacher's Guide ~ 10 or, 2×m2 =–1 m2 = –1 2 Now, equation of st. line RS is: y–y1 =m(x–x1 ), or, y – 8=– 1 2(x–6) x+2y =22 is the required equation. (c) P(2,4) and Q(–2,–4) Solution Let N be the mid point of RS Then the coordinates of S are 2–2 2 , 4–4 2 =(0,0) Slope of PQ (m1 )= y2 –y1 x2 –x1 = –4–4 –2–2 =2 Let MN be the perpendicular bisector of RS, then we get slope of MN is (m2 )=– 1 2 , ( m1 .m2 = –1 ) Now, equation of RS is y–y1 =m(x–x1 ), where m=m2 =– 1 2 or, y – 0=– 1 2(x–0) or, 2y=–x x+2y =0 is the required equation. 14 (a) In rhombus PQRS P(2,4) and R(8,10) are the opposite vertices. Find the equation of diagonal QS. Solution In rhombus PQRS the diagonals bisect each other at right angles. Let M be the mid point of the diagonals. Then the coordinates of M are 2+8 2 , 4+10 2 =(5,7) Slope of PR (m1 )= y2 –y1 x2 –x1 = 10–4 8–2 =1 Since PR QS, slope of QS (m2 )=– 1 ( m1 .m2 = –1 ) Now, equation of QS is y–y1 =m(x–x1 ), or, y – 7=–1(x–5) or, y–7 =–x+5 x+y =12 is the required equation. N S(–2,–4) M R(2,4) R(8,10) P(2,4) S Q M
Vedanta Optional Mathematics Teacher's Guide ~ 10 209 14(b) M(5,1) and P(–3,3) are two opposite vertices of square MNPQ. Find the equation of diagonal QN. Solution MNPQ is a square with opposite vertices M(5,1) and P(–3,3). In a square the diagonals bisect each other. Mid point of MP is 5–3 2 , 1+3 2 =(1,2) Slope of MP (m2 )= y2 –y1 x2 –x1 = 3–1 –3–5 =– 1 4 MP is perpendicular to NQ. Slope of QN (m2 )=4 Hence equation of QN passing through (1,2) and wiht slope 4 is given by, y–y1 =m(x–x1 ), or, y – 2=4(x–2) or, y – 2 =4x–8 4x – y =6 is the required equation. 15(a) Determine the equation of straight lines through (1,–4) that make an angle of 45˚ with the straight line 2x +3y+7=0. Solution Let MN be the given line with equation 2x+3y+7=0 Its slope is (m1 ) = – Coefficient of x Coefficient of y =– 2 3 Let PM and PN be two lines passing through P(1,–4) which make 45° with MN. Let m2 be the slope of PM or PN Then required equation are given by y – y1 =m(x–x1 ) or, y+4=m(x–1)............(i) Angle between the lines θ=45° Now, tanθ =± (m1 –m2 ) 1 +m1 m2 tan45º =± – 2 3 –m2 1+ –2 3 m2 or, 1= ± (2+3m2 ) 3–2m2 P(–3,3) M(5,1) (1,2) Q N P(1,–4) M N 45˚ 45˚ 2x+3y+7=0
210 Vedanta Optional Mathematics Teacher's Guide ~ 10 or, 3–2m2 =±(2+3m2 ) Taking positive sign, we get 3–2m2 = 2+3m2 or, –5m2 = –1 m2 = 1 5 Taking negative sign, we get 3–2m2 = 2+3m2 m2 = –5 When m2 = 1 5 ,then from equation(i),we get y +4 = 1 5 (x–1) or, 5y+20 = (x–1) x – 5y =21 When m2 =–5, then, from equation(ii),we get y +4 =–5(x–1) or, 5x +y =1 Hence the required equations are x – 5y =21 and 5x +y =1. (b) Find the equation of the straight lines passing through the point (3,2) and making angle of 45˚ with the line x–2y–3=0. Solution Let MN be the line with equation x–2y–3=0..........(i) slope is (m1 ) = – Coefficient of x Coefficient of y =– 1 –2 = 1 2 Let MP and NP be two lines passing through P(3,2) which make 45° with MN. and the slope m. Let the slope of MP or NP be m. Then required equation are given by y – y1 =m(x–x1 ) or, y–2=m(x–3)............(i) Angle between the lines θ=45° Now, tanθ =± (m1 –m2 ) 1 +m1 m2 , where m2 = m tan45º =± m – 1 2 1+ 1 2 m P(3,2) M N 45˚ 45˚
Vedanta Optional Mathematics Teacher's Guide ~ 10 211 or, 1= ± (1–2m) 2+m or, 2+m=±(1–2m) Taking positive sign, 2+m=1–2m or, 3m=-1 m= – 1 3 Taking negative sign, we get, 2+m=–1+2m m= 3 When m= – 1 3 , then from equation(i),we get y –2 = – 1 3 (x–3) or, 3y–6 =–x+3 ⸫ x +3y=9 When m=3, then from equation(i),we get y –2 =3(x–3) or, 3x–y =7 Hence the required equations are x + 3y =9 and 3x–y =7. (c)Find the equation of two lines passing through the point (1,–4) and making an angle of 45˚ with the lines 2x–7y+5=0. Solution Let MN be the line with equation 2x–7y+5=0 Its slope is (m1 ) = – Coefficient of x Coefficient of y =2 7 Let required lines be MP and NP passing through (1,–4) and the slope m. Let the slope of MP or NP be m. Then required equation is given by y – y1 =m(x–x1 ) or, y+4=m(x–1)............(i) Angle between the lines θ=45° Now, tanθ =± (m1 –m2 ) 1 +m1 m2 tan45º =± 2 7 –m 1+ 3 7 m P(1,–4) M N 45˚ 45˚ 2x–7y+5=0
212 Vedanta Optional Mathematics Teacher's Guide ~ 10 or, 1= ± (2–7m) 7+2m Taking positive sign, we get 7+2m=2–7m m= – 5 9 Taking negative sign, we get, 7+2m=–2+7m m= 9 5 When m= – 5 9 , from equation(i),we get y +4 = – 5 9 (x–1) or, 9y+36 =–5x+5 5x + 9y +31=0 When m= 9 5 , from equation(i),we get y +4 = 9 5 (x–1) or, 9x–5y =29 Hence the required equations are 5x + 9y +31 and 9x–5y =29. (d)Find the eqn. of straight line passing through the point (3,–2) and making an angle of 60˚ with the line 3 x+y–1=0. Solution Let the given line be QR with equation 3 x+y–1=0. Its slope is (m1 ) = – Coefficient of x Coefficient of y =– 3 Let QP and RP be two lines passing through P(3,–2) making angle 60° with QR. Let the slope of QP or RP be m. Then the required equation is y – y1 =m(x–x1 ) or, y+2=m(x–3)............(i) Now, tanθ =± m1 –m2 1 +m1 m2 tan60º =± (– 5 –m) 1 +(– 3 )m or, 3=± ( 3 +m) 1 – 3 m P(3,–2) Q R 60˚ 60˚ 3x+y–1=0
Vedanta Optional Mathematics Teacher's Guide ~ 10 213 Taking negative sign, we get 3–3m=– 3–m or, 2m=-2 3 m= 3 Taking positive sign, we get, 3–3m= 3+m ∴ m=0 When m= 3, then from equation(i), y +2 = 3(x–3) or, y+2 = 3x–3 3 or, 3x-y=3 3 +2 When m = 0, then from equation (i), y + 2 = 0 Hence the required equations are y+2=0 and 3x-y=3 3 +2. (e) Determine the value of m so that 3x–my–8=0 will make an angle of 45° with the line 3x+5y–17=0. Solution :Let MN and MP be the given lines with angle 45° between them. Equation MN : 3x–5y–8=0. slope of MN (m1 ) =– 3 –m = 3 m Equation MN : 3x+5y–17=0 slope of MP(m2)= – 3 5 Now, tanθ =± 3 m + 3 5 1+ 3 m – 3 5 or, tan45°=± 3 m + 3 5 1+ 3 m . – 3 5 or, 1 =± 15 +3m 5m–9 or, 5m–9=±(15+3m) Taking positive sign, we get 5m–9=15+3m or, 2m=24 m = 12 P M N 45˚ 3x–my–8=0 3x+5y–17=0
214 Vedanta Optional Mathematics Teacher's Guide ~ 10 Taking negative sign, we get, 5m–9=–15–3m or, 8m=–6 ⸫ m=12 or – 3 4 . 16(a). Find the equation of the sides of an equilateral triangle whose vertex is (1,2) and base is y =0. Solution Let ABC be an equilateral triangle with vertex A(1,2) with base y=0 there . ABC= BCA= CAB=60° slope of BA=tan60°= 3 slope at CA=tan120° =– 3 Now, for equation of BA let, (x1 ,y1 )=(1,2) (y–2)= 3 (x–1) or, 3 x –y+2– 3 =0 ⸫ 3 x–y+2– 3 =0 Again, for equation of CA (y – 2)=– 3 (x – 1) or, y –2 =– 3 x– 3 ⸫ 3 x + y+ 3 –2=0 Hence the required equations are 3 x–y+2– 3 =0 and 3 x + y+ 3 –2=0. (b). Find the equation of the sides of right angled isoceles triangle whose vertex is at (–2,–3) and base is x =0. Solution: Let P(–2,–3) be the vertex of right angled isoceles triangle PQR with P=90° For side PR, its slope (m1 )=tan45° Equation of PR is y–y1 =m(x–x1 ) or, y +3=1(x+2) x – y=1 Again, for equation of side QP, slope (m)=tan135°=–1 Equation of QP is y–y1 =m(x–x1 ) or, y +3=–1(x+2) or, x+y+5=0 ⸫ Hence the required equation are x – y=1 and x+y+5=0. A(1,2) C x x' y B 60˚ 60˚ 60˚ 120˚ 0 y=0 P(–2,–3) x=0 R Q x' x y y' 135˚ 45˚ O
Vedanta Optional Mathematics Teacher's Guide ~ 10 215 (c). Find the equation of the sides of right angled isoceles triangle whose vertex is at (2,4) and equation of base x =0. Solution Let ∆MNP be an isoceles right angled triangle with p=90° For side NP, slope (m1 )=tan45°=1 Now, equation of NP is y–y1 =m(x–x1 ), or, y – 4=1(x–2) x–y+2=0 For side MP, slope (m2 )=tan135° = –1 Equation of MP is y–y1 =m(x–x1 ) or, y – 4=–1(x–2) x+y=6 Hence the required equation are x +y+2=0 and x+y=6. Questions for practice 1. The line px+3y+5=0 is perpendicular to the line joining the points (4,3) and (6,–3), find the values of P. 2. Find the equation of a straight line passing through the point of intersection of the straight lines x–y=7 and x+y=15 and parallel to the line 4x+3y=10. 3. Find the equation of the line segment PQ which passes through the point (3,4) and the mid point of line segment joining M(–4,–5) and N(7,8). 4. If the line x a + y b = 1 passes through the point of intersection of the lines x+y=3 and 2x–3y=1 and is parallel to the line x–y=6, then find the values of a and b. 5. Find the equation of perpendicular bisector of line segment joining M(–4,–5) and N(8,9). Pair of straight lines Estimated Teaching periods : 8 1. Objectives S. N. Level Objectives 1. Knowledge(K) To define equation of pair of lines. To define homogeneous equation. To tell general equation of second degree. To tell formula to find angle between two lines represented by ax2 +2hxy+by2 =0 To state conditions for coincidence and perpendicularity. P(2,4) x=0 x' x y M N y
216 Vedanta Optional Mathematics Teacher's Guide ~ 10 2. Understanding(U) To find a single equation of a pair of lines. To find pair of lines of given homogeneous equation in x and y. To identify given homogeneous equation represents a pair of perpendicular or coincident lines. 3. Application(A) To find angle between two lines represented by general equation of second degree. To find separate equation of two lines represented by general eqn. of second degree. 4. Higher Ability (HA) To derive formula to find angle between two lines represented by ax2 +2hxy+by2 =0 and find condition for coincidence and perpendicularity To show ax2 +2hxy+by2 =0 represents a pair of lines through origin. 2. Required teaching materials : Chart paper with figure of angle between two lines represented by ax2 +2hxy+by2 =0. 3. Teaching Learning Activities: – Take two equations like x+2y=0 and x–2y=0 and multiply them to get x2 –4y2 =0, discuss the conclusion. – Find two separate equations represented by x2 +8xy+12y2 =0, discuss the conclusion. – Define general equation of second degree in x and y . – Define a homogeneous equation with examples. – Show that ax2 +2hxy+by2 =0. represents a pair of lines through the origin. – Discuss how to find the angle between the lines represented by ax2 +2hxy+by2 =0. ? – Discuss the conditions for perpendicularity and coincidence of the lines represented by ax2 +2hxy+by2 =0. – Do some problems from exercise to guide the students. Note: i) The equation ax2 +2hxy+by2 =0 always represents a pair of lines through the origin in the form of y=m1 x and y=m2 x. 2. The angle between the lines represented by ax2 +2hxy+by2 =0 is =tan-1 a+b h2 ± 2 –ab 3. The ax2 +2hxy+by2 =0 represents a pair of lines, then i) Condition for coincidence of the lines h2 =ab ii) Condition for perpendicularity/orthogonality a+b=0 4. The roots of quadratic equation ax2 +bx+c=0 are x= 2a –b± b2 –4ac
Vedanta Optional Mathematics Teacher's Guide ~ 10 217 Some solved problems 1. Find the single equation represented by the pair of straight lines. 3x+4y=0 and 2x–3y=0 Solution Given equations of lines are 3x+4y=0 and 2x–3y=0 the single equation represented by above equations is (3x+4y)(2x–3y)=0 or, 6x2 –9xy+8xy–12y2 =0 or, 6x2 –xy–12y2 =0 6x2 –xy–12y2 =0 is the required equation. 2. Find the separate equations of of straight lines represented by the following equations. x2 + y2 –2xy+2x–2y =0 Solution Here,x2 + y2 –2xy+2x–2y =0 or, (x2 –2xy+y2 )+2(x–y) =0 or, (x–y)2 +2(x–y) =0 or, (x–y)(x–y+2) =0 Hence x–y =0 and x–y+2 =0 are the required equations. 3. Show that 6x2 –5xy–6y2 =0 represents a pair of lines. Solution Here, 6x2 –5xy–6y2 =0 or, 6x2 –9xy+4xy–6y2 =0 or, 3x(2x–3y)+2y(2x–3y)=0 or, (2x–3y)(3x–2y)=0 Either 2x–3y=0 or 3x–2y=0 , each of which are equation of straight lines. 4. Determine the lines represented by each of the given equations. (a) x2 +2xy+y2 –2x–2y–15=0 Solution Here, x2 +2xy+y2 –2x–2y–15=0 or, (x+y)2 –2(x+y)–15=0 or, (x2 +y)2 –5(x+y)+3(x+y)–15=0 or, (x+y) (x+y–5)+3(x+y–5)=0 or, (x+y–5) (x+y+3)=0 Either, or, (x+y–5)=0 or (x+y+3)=0
218 Vedanta Optional Mathematics Teacher's Guide ~ 10 Each of which represents a straight line Hence the given equation represents a pair of lines. (b) 2x2 +xy–3y2 +2y–8=0 Solution Here, 2x2 +xy–3y2 +2y–8=0 or, 2x2 +xy+(–3y2 +10y–8)=0 which is in the form of ax2 +bx+c=0, where, a=2, b=y and c=-3y2 +10y–8=0 By using formula, x= 2a –b± b2 –4ac = 2.2 –y± y2 –4.2.(-3y2 +10y–8) = 2.2 –y± y2 +24y2 –80y+64 = 4 –y± (5y–8)2 Taking positive sign, we get 4x=–y+5y–8 or, 4x–4y+8=0 or, x–y+2=0 Taking negative sign, we get 4x=–y–5y+8 or, 4x+6y–8=0 or, 2x+3y–4=0 Hence the required equation are x–y+2=0 and 2x+3y =4 (c) x2 +9y2 +6xy+4x+12y–5=0 Solution Here, x2 +6xy + 9y2 +4x+12y–5=0 or, x2 +2.x.3y + (3y)2 +4(x+3y)–5=0 or, (x+3y)2 +4(x+3y)–5=0 or, (x+3y)2 +5(x+3y)–(x+3y)–5=0 or, (x+3y)(x+3y+5)–1(x+3y+5)=0 or, (x+3y+5)(x+3y–1)=0 Hence the required equation are x+3y+5=0 and (x+3y–1)=0
Vedanta Optional Mathematics Teacher's Guide ~ 10 219 5. Prove that two lines represented by the following equations are perpendicular to each other. 9x2 –13xy–9y2 +2x–3y+7=0 Solution Here, 9x2 –13xy–9y2 +2x–3y+7=0 Comparing at with ax2 –2hxy+by2 +2gx+2fy+c=0 we get, a=9, h=– 13 2 , b =-9, g=1, f= – 3 2 , c=7 condition for perpendicularity a+b=0 Here, a+b=9–9=0 a+b=0 Hence the lines represented by given equation are perpendicular to each other. 6. Show that the given equations represents a pair of coincident lines. 9x2 –6xy+y2 = 0 Solution Here, 9x2 –6xy+y2 = 0 Comparing it with ax2 –2hxy+by2 =0 we get, a=9, b =1, g=1, h=–3 condition for coincident, h2 =ab ie. (–3)2 =9.1 9=9(true) proved. 7. Find the value of λ when the lines represented by each of the following equations are coincident. (10λ–1)x2 +(5λ+3)xy+(λ–1)y2 =0 Solution: Comparing given equation (10λ–1)x2 +(5λ+3)xy+(λ–1)y2 =0 with ax2 +2hxy+by2 +2gx+2fy+c=0 , we get, a= 10λ–1, h= 5λ+3 2 ,b= λ–1 Now, condition for coincident h2 =ab or, 5λ+3 2 2 = (10λ–1). (λ–1) or, 25λ2 +30λ+9 4 = 10λ2 –10λ–λ+1 or, 25λ2 +30λ+9= 40λ2 –44λ +4 or, 15λ2 –74λ–5=0 or, 15λ2 –75λ+λ–5=0 or, 15λ(λ–5) +1(λ–5)=0
220 Vedanta Optional Mathematics Teacher's Guide ~ 10 or, (λ–5)(15λ+1)=0 Either λ–5=0 or λ=5 or, 15λ+1=0 or, λ= – 1 15 λ=5,– 1 15 8. Find the value of λ when the give equation represents a pair of lines perpendicular to each other. (3λ+4)x2 –48xy–λ2 y2 =0 Solution Comparing the given equation with ax2 +2hxy+by2 =0, we get a= 3λ+4, h=–24 ,b=–λ2 Condition for perpendicularity, a+b=0 or, 3λ+4–λ2 = 0 or,λ2 –3λ–4= 0 or,λ2 – 4λ+λ – 4=0 or, λ (λ– 4)+1(λ – 4)=0 or, (λ– 4)(λ +1)=0 Either , λ– 4=0 or λ=4 or, λ+1=0 or λ=–1 λ=4,–1 9. Find the angle between the following pair of lines. (a) x2 +9xy+14y2 =0 Solution Here, x2 +9xy+14y2 =0 comparing it with ax2 +2hxy+by2 =0, we get a=1, h= 9 2 , b=14 Let θ be the angle between them. tanθ= a+b h2 ±2 –ab = 1+14 81 4 ±2 –14 = ±2 2×15 81–56 =± 5 15
Vedanta Optional Mathematics Teacher's Guide ~ 10 221 =± 13 taking positive sign, tan θ = 13 θ=71.57 ° taking negative sign, tanθ= – 13 θ=108.43 ° (b) 2x 2 +7xy+3y 2 =0 Solution Comparing it with ax 2 +2hxy+by 2 =0, we get a=2, h= 72 , b=3 Let θ be the angle between the pair of lines, we get, tanθ= a+bh2 ±2 –ab = 2+3 494 ±2 –2.3 = ±2 5 2.5 =±1 taking positive sign, we get, tan θ=1 or, tan θ=tan45 ° θ=45 ° taking negative sign, we get tanθ= – 1 or, tan θ=tan135° θ=135 ° Hence, θ=45° ,135°. (c) x 2 –7xy+y 2 =0 Solution Comparing it with ax 2 +2hxy+by 2 =0, we get a=1, h=– 72 , b=1 Let θ be the angle between the pair of lines, we get, tanθ= a+bh2 ±2 –ab = 1+1 494 ±2 –1
222 Vedanta Optional Mathematics Teacher's Guide ~ 10 = ± 2 3 5 taking positive sign, we get, tanθ= 2 3 5 θ=73.4° taking negative sign, we get tanθ= – 2 3 5 or, tanθ=tan135° θ=106.6° Hence, θ=73.4° ,106.6°. (d) 9x2 –13xy–9y2 +2x–3y +7=0 Solution comparing it with ax2 +2hxy+by2 +2gx+2fy+c=0, we get a=9, h=– 13 2 , b=–9, g=1, f=–– 3 2 , c=7 Let θ be the angle between the pair of lines, we get, tanθ= a+b h2 ±2 –ab = 9–9 169 ±2 4 +81 = ∞ =tan90° θ=90° (e) x2 +6xy+9y2 +4x+12y –5=0 Solution comparing it with ax2 +2hxy+by2 +2gx+2fy+c=0, we get a=1, h=3, b=9, g=2, f= 6, c=–5 Let θ be the angle between the pair of lines, tanθ= a+b h2 ±2 –ab = 1+9 ±2 9–9 = 0 or, tanθ=tan0° θ=0° 10. Find the angle between the following pair of lines. (a) x2 –2xycot –y2 =0
Vedanta Optional Mathematics Teacher's Guide ~ 10 223 Solution comparing given equation with ax2 +2hxy+by2 =0, we get a=1, h=–cot , b=–1 Let θ be the angle between the pair of lines, tanθ= a+b h2 ±2 –ab = 1–1 cot ±2 2 +1 = ∞ or, tanθ=tan90° θ=90° (b) x2 +2xycosec +y2 =0 Solution Comparing given equation with ax2 +2hxy+by2 =0, we get a=1, h=cosec , b=1 Let θ be the angle between the pair of lines, tanθ= a+b h2 ±2 –ab = 1+1 cosec2 ±2 –1 =±cot or, tanθ=tan(90°± ) θ=90°± (c) y2 +2xycot –x2 =0 Solution The given equation can be written as y2 +2xycot –x2 =0 Comparing given equation with ax2 +2hxy+by2 =0, we get a=1, h=–cot , b=–1 Let θ be the angle between the pair of lines, tanθ= a+b h2 ±2 –ab = 1–1 cot ±2 2 +1 =∞ θ=90°
224 Vedanta Optional Mathematics Teacher's Guide ~ 10 11. Find the separate equation of two lines represented by the following equations. (a) x2 +2xycot –y2 =0 Solution Here, x2 +2xycot –y2 =0 or, x2 – 2xycot – y2 (cosec2 –cot2 )=0 or, x2 + 2xycot + y2 cot2 – y2 cosec2 =0 or, (x+ycot ) 2 – (ycosec ) 2 =0 or, (x+ycot + ycosec )(x+ycot – ycosec )=0 Hence the required separate equations are x + y(cot + cosec )=0 and x – y(cot – cosec )=0 (b) x2 + 2xysec + y2 =0 Solution Here, x2 + 2xysec + y2 =0 or, x2 + 2xysec + y2 (sec2 –tan2 )=0 or, x2 + 2xysec + y2 sec2 – y2 tan2 =0 or, (x+ysec ) 2 – (ytan ) 2 =0 or, (x+ysec + ytan )(x+ysec – ytan )=0 Hence the required separate equations are x + y(sec + tan )=0 and x + y(sec – tan )=0 (c) x2 +2xycosec + y2 =0 Solution Here, x2 +2xycosec + y2 =0 or, x2 + 2xycosec + y2 (cosec2 –cot2 )=0 or, (x2 + 2xycosec + y2 cosec ) – y2 cot2 =0 or, (x+ycosec ) 2 – (ycot ) 2 =0 or, (x+ycosec + ycot )(x+ycosec – ycot )=0 Hence the required separate equations are x+y(cosec + cot ) = 0 and x+y(cosec – cot )=0 (d) x2 – 2xycot2 –y2 =0 Solution Here, x2 – 2xycot2 –y2 =0 or, x2 – 2xycot2 – y2 (cosec2 2 –cot2 2 )=0 or, x2 – 2xycot2 + y2 cot2 2 – y2 cosec2 2 =0 or, (x – ycot2 ) 2 – (ycosec2 ) 2 =0
Vedanta Optional Mathematics Teacher's Guide ~ 10 225 or, (x – ycot2 + ycosec2 )(x – ycot2 – ycosec2 )=0 or, {x – y(cot2 – cosec2 )}{x – y(cot2 + ycosec2 )}=0 Hence the required separate equations are x – y(cot2 – cosec2 )=0 and x – y(cot2 + cosec2 )=0 14(a). Find the equation of two lines represented 2x2 + 7xy + 3y2 =0. Find the point of intersection . Also find the angle between them. Solution Here, 2x2 + 7xy + 3y2 =0 or, 2x2 + 6xy + xy + 3y2 =0 or, 2x(x + 3y) + y(x+3y) =0 or, (x + 3y)(2x + y)=0 Hence the equation of two lines are x + 3y = 0.................(i) 2x + y = 0...............(ii) solving equations (i) and (ii) , we get x=0 and y = 0 (0,0) is the point of intersection of lines (i) and (ii) To find the angle between the lines. comparing the given equation 2x2 + 7xy + 3y2 =0 we get, a= 2, h = 7 2 and b =3 Let θ be the angle between the lines tanθ = a+b h2 ±2 –ab = 1+1 49 4 ±2 –2.3 = ± 5 2 2 . 5 = ±1 taking positive sign, we get, tanθ= 1 = tan45° θ=45° taking negative sign, we get ttanθ= –1 = tan135° θ=45° θ= 45° or 135°
226 Vedanta Optional Mathematics Teacher's Guide ~ 10 14(b). Find the separate equation of two lines represented x2 – 5xy + 4y2 =0. Also find the angle between them and their point of intersection. Solution Here, x2 – 5xy + 4y2 =0 or, x2 – 5xy + 4y2 =0 or, x(x – 4y) – y(x – 4y) =0 or, (x – 4y)(x – y)=0 Hence the equation of two lines are x – 4y = 0.................(i) x – y = 0...............(ii) solving equations (i) and (ii) , we get (x, y) = (0,0) comparing the given equation x2 – 5xy + 4y2 =0, we get, a= 1, h =– 5 2 and b = 4 Let θ be the angle between the lines tanθ = a+b h2 ±2 –ab = 1+4 25 4 ±2 –1.4 = ± 5 2 θ=tan–1 ± 3 5 , θ=31°,149° (c) Find the separate equations of two lines represented by the equation x2– 2xycosec + y2=0 . Also find the angle between them. Solution Here, x2 – 2xycosec + y2 =0 or, x2 – 2xycosec + y2 (cosec2 –cot2 )=0 or, (x2 – 2xycosec + y2 cosec ) – y2 cot2 =0 or, (x – ycosec ) 2 – (ycot ) 2 =0 or, (x – ycosec + ycot )(x – ycosec – ycot )=0 Either x – y(cosec – cot ) =0...............(i) x – y(cosec + ycot )=0...............(ii) Which are the required equation of straight lines
Vedanta Optional Mathematics Teacher's Guide ~ 10 227 Comparing the given equation with ax2 +2hxy + by2 =0, we get, a= 1, h =–cosec , b = 1 Let θ be the angle between the lines tanθ = a+b h2 ±2 –ab = 1 + 1 cosec2 ±2 – 1 = ±cot = ±tan(90˚ ± ) = tan(90˚ ± ) θ = 90˚ ± 15(a). Find the pair of lines parallel to the lines x2 – 3xy + 2y2 = 0 and passing through the origin. Solution Here, x2 – 3xy + 2y2 = 0 or, x2 – 2xy – xy + 2y2 = 0 or, x(x – 2y) –y(x – 2y) =0 or, (x – 2y) (x – y) = 0 x – 2y = 0..........(i) and x – y = 0 ..........(ii) are the required equations. Equations of the lines parallel to (i) and (ii) and passing through the origin are, x – 2y = 0 and x – y = 0 (b). Find the pair of lines parallel to the 2x2 – 7xy + 5y2 = 0 and passing through the point (1,2). Solution Here, 2x2 – 7xy + 5y2 or, x2 – 5xy – 2xy + 5y2 = 0 or, x(2x – 5y) –y(2x – 5y) =0 or, (2x – 5y) (x – y) = 0 2x – 5y = 0..........(i) and x – y = 0 ..........(ii) Equations of the lines parallel to (i) and (ii) are, 2x – 5y + k1 = 0 ...........(iii) and x – y + k2 = 0 ................(iv) The lines (i) and (ii) passes through the point (1,2), we get, 2 × 1 – 5 × 2 + k1 = 0 ⇒ k1 =8 and 1 – 2 + k2 = 0 ⇒ k2 =1 put the values of k1 and k2 in eqn. (iii) and (iv), we get, 2x – y + 8 = 0 and x – y + 1 = 0.
228 Vedanta Optional Mathematics Teacher's Guide ~ 10 16(a). Find the pair of lines perpendicular to the lines x2 – 5xy + 4y2 = 0 and passing through the origin. Solution Here, x2 – 5xy + 4y2 = 0 or, x2 – 4xy – xy + 4y2 = 0 or, x(x – 4y) –y(x – 4y) =0 or, (x – 4y) (x – y) = 0 Either, x – 4y = 0..........(i) or, x – y = 0 ..........(ii) From equations (i), equation of the lines perpendicular to it and passing through the origin are, 4x + y = 0 Also, from equation (ii), equation of line perpendicular to it is x + y = 0 Required pair of lines are 4x + y = 0 and x + y = 0. (b). Find the equation of two straight lines which pass through the point (2,3) and perpendicular to the lines x2 – 6xy + 8y2 = 0. Solution Here, x2 – 6xy + 8y2 = 0 or, x2 – 4xy – 2xy + 8y2 = 0 or, x(x – 4y) – 2y(x – 4y) =0 or, (x – 4y) (x – 2y) = 0 Either x – 4y = 0..........(i) or, x – 2y = 0 ..........(ii) Equations of the lines perpendicular to above lines are, 4x + y + k1 = 0 ...........(iii) and 2x – y + k2 = 0 ................(iv) Both of above lines passes through the point (2,3), we get, 4 × 2 + 3 + k1 = 0 →k1 = –11 and 2 + 2×3 + k2 = 0 →k2 = –7 put the values of k1 and k2 in eqn. (iii) and (iv), we get, 4x + y –11 = 0 and 2x+y – 7 = 0 which are the required equation of the lines. 17 (a). Find the two separate equations when the lines represented by kx2 + 8xy – 3y2 = 0 are perpendicular to each other . Solution Here, kx2 + 8xy – 3y2 = 0 ...................(i) Comparing it with ax2 + 2hxy + by2 = 0, we get, a = k, b = –3, h = 4
Vedanta Optional Mathematics Teacher's Guide ~ 10 229 Conditions for perpendicularity is a + b = 0 or, k – 3 = 0 or, k ⇒ 3 put the value of k in eqn.(i), we get 3x2 + 8xy – 3y2 = 0 or, 3x2 + 9xy – xy – 3y2 = 0 or, 3x(x + 3y) – y(x + 3y) =0 or, (x + 3y) (3x – y) = 0 x + 3y = 0 and 3x – y = 0 are the required equation of the lines. (b). Find the two separate equations when the lines represented by 6x2 + 5xy – ky2 = 0 are perpendicular. Solution Here, 6x2 + 5xy – ky2 = 0 ...................(i) Comparing it with ax2 + 2hxy + by2 = 0, we get, a = 6, b = –k, h = 5 2 conditions for perpendicularity is a + b = 0 or, 6 – k = 0 k = 6 put the value of k in eqn.(i), we get 6x2 + 5xy – 6y2 = 0 or, 6x2 + 9xy – 4xy – 6y2 = 0 or, 3x(2x + 3y) – 2y(2x + 3y) =0 or, (2x + 3y) (2x – 2y) = 0 2x + 3y = 0 and 2x – 2y = 0 are the required equation of the lines. 18(a). Show that the pair of lines 3x2 – 2xy – y2 = 0 are parallel to the lines 3x2 – 2xy – y2 –5x + y + 2 =0 Solution Here, 3x2 – 2xy – y2 = 0 or, 3x2 – 3xy + xy – y2 = 0 or, 3x(x – y) + y(x – y) =0 or, (x – y) (3x + y) = 0 Either, x – y = 0 ...........(i) 3x + y = 0 ..........(ii) Also, 3x2 – 2xy – y2 –5x + y + 2 =0 3x2 – (2y + 5)x + (y – y2 + 2) =0 which is in the form of ax2 + bx + c = 0
230 Vedanta Optional Mathematics Teacher's Guide ~ 10 where a = 3, b = – (2y + 5), c = y – y2 + 2 Comparing it with ax2 + 2hxy + by2 = 0, we get, x= 2a –b± b2 –4ac = 2 . 3 (2y + 5)± (2y + 5)2 – 4 . 3.(y – y2 + 2) = 6 (2y + 5)± 4y2 +20y +25 –12y +12y2 – 12) = 6 2y± 16y2 +8y + 1 = 6 2y± (4y + 1)2 = 2y ± (4y + 1) Taking positive sign, we get 6x = 2y+4y + 1 or, 6x – 6y – 1=0 ........... (iii) or, x–y+2=0 Taking negative sign, we get 6x = 2y – 4y – 1 or, 6x + 2y + 1=0 ........... (iv) From eqn.(i), slope (m1 ) = 1 From equation (iii), slope (m3 ) = 1 m1 = m3 The lines (i) and (iii) are parallel to each other. Again, from equation (ii), slope(m2 ) = –3 from equation (iv), slope (m4 ) = – 6 2 =–3 m2 = m4 The lines (ii) and (iv) are parallel to each other . proved 18(b). Show that the pair of lines 4x2 – 9y2 = 0 and 9x2 – 4y2 = 0 are perpendicular to each other. Solution Here, 4x2 – 9y2 = 0 or, (2x – 3y) (2x + 3y) = 0 Either, 2x – 3y = 0 ...........(i) or, 2x + 3y = 0 ..........(ii) Again ,9x2 – 4y2 = 0
Vedanta Optional Mathematics Teacher's Guide ~ 10 231 (3x + 2y) (3x – 2y) = 0 Either, 3x + 2y = 0 ...........(iii) or, 3x – 2y = 0 ..........(iv) From equation (i), slope (m1 ) = 2 3 From equation (iii), slope (m3 ) = – 3 2 m1 . m3 = 2 3 – 3 2 = –1 The lines (i) and (iii) are perpendicular to each other. Again, from equation (ii), slope(m2 ) =– 2 3 from equation (iv), slope (m4 ) = 3 2 Now, m2 . m4 = – 2 3 3 2 = –1 The lines (ii) and (iv) are perpendicular to each other . proved Questions for practice 1. Find the single equation representing the pair of lines y = x and y = –x. 2. Find the two separate equations represented by i) 2x2 + 7xy + 3y2 = 0 ii) x2 + 2xysecθ + y2 = 0 3. Determine the two straight lines represented by 6x2 – xy – 12y2 –8x + 29y –14 =0 4. Find the equation to the straight lines through the origin and at right angles to the lines x2 – 5xy + 4y2 = 0 5. Find the value of k when the pair of lines represented by (k + 2)x2 +8xy + 4y2 = 0 are coincident. 6. Show that the angle between a pair of lines represented by 2x2 – 7xy + 3y2 + 2x – 6y=0 is 60°. 7. Find the two separate equations of straight lines passing through the point (1,0) and parallel to the lines represented by the equation x2 + 3xy + 2y2 = 0 8. Find the equation of two lines which pass through the point (3, –1) and perpendicular to the pair x2 – xy – 2y2 = 0. 9. Find the value of p so that the two lines represented by the equation (p + 1)x2 – 12xy + 9y2 = 0 are coincident. 10. Prove that the lines represented by x2 – 7xy + 12y2 = 0 are perpendicular to the lines represented by 12x2 + 7xy + y2 = 0
232 Vedanta Optional Mathematics Teacher's Guide ~ 10 Conic Section and Circle Estimated Teaching periods : 10 Hours 1. Teaching Objectives S,N. Level Objectives 1. Knowledge(K) To define terms vertex axis, generator of a cone. To define ellipse, parabola, hyperbola. 2. Understanding (U) To identify types of conic sections from given figures. – To derive equation of circle x2 +y2 = a2 and (x –h)2 + (y – k)2 = r2 by using distance formula. 3. Application (A) To use equation of circles (x – h)2 + (y – k)2 = r2 (x – x1 )(x – x2 ) + (y – y1 )(y – y2 ) = 0 to find equation of circles. 4. Higher Ability (HA) To solve verbal problems related to circle – eqn of circle passing through three or more given points. To show given four points concyclic. 2. Required teaching materials – diagrams of conic sections. – graph papers. 3. Teaching strategies – Discuss different types of conic sections – Circle, parabola, ellipse, hyperbola by using plane figures of intersection of a plane and a cone. – Discuss to derive the following equations of circles x2 + y2 = a2 – (x – h)2 +(y – k)2 = r2 , (x – h)2 +(y – h)2 = h2 , (x – k)2 +(y – k)2 = k2 – (x – x1 )(x – x2 ) + (y – y1 )(y – y2 ) = 0 – x2 + y2 +2gx + 2fy + c = 0 To each of formula, illustrated examples are to be given. Notes : Equations of circles in different forms. i) Equation of circle with centre at the origin 0(0,0) and radius r : x2 + y2 = a2 ii) Equation of circle with centre at (h, k) and radius r; (x – h)2 + (y – k)2 = r2 iii) Equation of circles touching x – axis ie. r = k, (x – k)2 +(y – k)2 = r2 iv) Equation of a circle touching y–axis, ie. r = h: (x – h)2 +(y – k)2 = r2 v) Equation of a circle touching both axis : h = k = r (x – h)2 +(y – h)2 = h2 ,
Vedanta Optional Mathematics Teacher's Guide ~ 10 233 vi) Equation of a circle in a diameter form (x – x1 )(x – x2 ) + (y – y1 )(y – y2 ) = 0 vii) General Equation of circle : x2 + y2 +2gx + 2fy + c = 0 where center (h,k) =(–g, –f) radius (r) = g2 + f2 – c If circle x2 + y2 +2gx + 2fy + c = 0 passes through origin, then c = 0, equation of circle is x2 + y2 +2gx + 2fy = 0. viii) The point of intersection of two diameters is the center of the circle. Some solved problems 1. Write radius and centre of circle x2 + y2 + 2gx + 2fy + c = 0 Solution Equation of circle is x2 + y2 + 2gx + 2fy + c = 0 Radius (r) = g2 + f2 – c centre (h,k) =(–g, –f) 2. Find the equation of circle with centre (–2, –3) and radius 6 units. Solution Here, centre (h,k)=(–2, –3) radius (r) = 6 units Equations of circle is (x – h)2 +(y – k)2 = r2 ie. (x + 2)2 +(y + 3)2 = 62 or, x2 + 2.x.2 + 22 +y2 +2.y.3 +32 =36 or, x2 + y2 + 4x +6y =36 3. Find the equation of circle whose end of diameters are (4,5) and (–2, –3). Solution Here, centre ( x1 , x2 ) = (4,5) ,(y1 , y2 ) = (–2, –3) Equations of the required circle is (x – x1 )(x – x2 ) + (y – y1 )(y – y2 ) = 0 ie. (x – 4)(x + 2) + (y – 5)(y + 3) = 0 or, x2 – 2x – 8 + y2 – 2y – 15 = 0 ie. x2 + y2 – 2x –2y =23 x2 + y2 – 2x –2y = 23 is the required equation. 4. Find the centre and radius of the circles. (a) x2 + y2 + 6x + 4y –12 = 0
234 Vedanta Optional Mathematics Teacher's Guide ~ 10 Solution Here, x2 + y2 + 6x + 4y –12 = 0 comparing it with x2 + y2 + 2gx + 2fy + c = 0 g = 3, f = 2, c = –12 centre (–g, –f) = (–3, –2) Radius (r) = g2 + f2 – c = 9 + 4 – 12 = 5 = 5 units. (b) 9x2 + 9y2 – 36x + 6y = 107 Solution Here, 9x2 + 9y2 – 36x + 6y = 107 dividing both sides by 9, we get, x2 + y2 – 4x + 2y 3 = 107 9 comparing it with x2 + y2 + 2gx + 2fy + c = 0 we get, g = –2, f = 1 3 , c = – 107 9 centre (–g, –f) = 2,– 1 3 Radius (r) = g2 + f2 – c = 22 + – 1 3 2 + 107 9 = 4 + 1 9 + 107 9 = 36 + 1 + 107 9 = 144 9 = 16 = 4 (c) x2 + y2 – 2axcosθ – 2aysinθ = 0 Solution Here, x2 + y2 – 2ax cosθ – 2ay sinθ = 0 comparing it with x2 + y2 + 2gx + 2fy + c = 0 g = –acosθ , h =–asinθ, we get
Vedanta Optional Mathematics Teacher's Guide ~ 10 235 centre (–g,–f) = (asinθ, asinθ) Radius (r) = g2 + f2 – c = (–acosθ)2 +(–asinθ) 2 = a2 cos2 θ + a2 sin2 θ = a2 (sin2 θ +cos2 θ) = a units. 5. Find the equation of circle whose centre is (4,5) and touches x – axis. Solution Here centre (h,k) = (4,5) The required circle touches x – axis at (4,0) and radius (r) = k=5 Now, the equation of the circle is, (x – h)2 +(y – k)2 = k2 ie. (x – 4)2 +(y – 5)2 = 52 x2 – 8x + 16 + y2 –10y +25 = 25 x2 + y2 –8x – 10y + 16 = 0 is the required equation. 6. Find the equation of a circle whose centre is (4,–1) and passing through (–2,–3) . Solution Distance between the centre (4,–1) and point (–2,–3) is the radius of the circle. radius (r) = distance between C(4,–1) and P(–2,–3) = (–2–4)2 +(–3+1)2 = 36 + 4 = 40 =2 10 units. Now, the equation of the circle is, (x – h)2 +(y – k)2 = r2 ie. (x – 4)2 +(y + 1)2 = 40 x2 – 8x + 16 + y2 + 2y + 1 = 40 x2 + y2 –8x + 2y= 23 is the required equation of the circle. 7. Find the equation of a circle whose centre is the p oint of intersection of x + 2y – 1 = 0 and 2x – y – 7 = 0 and passing through the point (6,4) . Solution Given equation of lines are, O (4,0) C(4,5) x' x y y' P(–2,–3) C(4,–1)
236 Vedanta Optional Mathematics Teacher's Guide ~ 10 x + 2y – 1 = 0 .........(i) 2x – y – 7 = 0 ..........(ii) Solving equations (i) and (ii) we get, (x, y) = (3, –1). The point of intersection of the lines (i) and (ii) is the centre of the circle. centre (h, k) = (3, –1) The point (6, 4) is on the circumference of the circle. radius (r) = distance between (3, –1) and a point on circumference. = (x2 – x1 ) 2 +(y2 – y1 ) 2 = (6 – 3)2 +(4 + 1)2 = 32 + 52 = 9 + 25 = 34 units. Hence the equation of circle is, (x – h)2 +(y – k)2 = r2 ie. (x – 3)2 +(y + 1)2 = 34 x2 + y2 – 6x + 2y= 24 is the required equation of the circle. 7. Determine the points of intersections of a straight line and the circle x + y = 3, x2 + y2 – 2x – 3 = 0. Also find the length of the intercepts(chord). Solution Given equation of lines are, x + y = 3........(i)(a line) x2 + y2 – 2x – 3 = 0 ..........(ii)(a circle) Solving equations (i) and (ii) we get, the point of intersection of the line and the circle. From equation (i), y = 3 – x ..........(iii) put the value of y in equation (ii), we get x2 + (3 – x)2 –2x –3 = 0 or, x2 + 9 – 6x +x2 –2x –3 = 0 or, 2x2 – 8x + 6 = 0 or, x2 – 4x + 3 = 0 or, x2 – 3x –x + 3 = 0 (4,0) (3,–1) 2x–y–7=0 x–2y–1=0 (6,4) r P Q x+y=3
Vedanta Optional Mathematics Teacher's Guide ~ 10 237 or, (x – 3)(x – 1) =0 x = 1, 3 From equation(i), when x = 1, then y = 2 when x = 3, then y = 0 Hence the required points of intersection are (1,2) and (3,0) Also, the length of chord, PQ = (3 – 1)2 +(0 – 2)2 = 4 + 4 = 8 = 2 2 units. 9. Find the centre and radius of circle passing through the points P(2, –1), Q(2,3) and R(4,1). Also find the equation of the circle. Solution Let the equation of required circle be (x – h)2 +(y – k)2 = r2 ...............(i) where centre = c(h,k), radius = r The circle (i) passes through the points P(2, –1), Q(2,3) and R(4,1). Now, CP =CQ =CR ie. CP2 = CQ2 =CR2 Taking CP2 = CQ2 ie. (2 – h)2 +(–1 – k)2 = (2 – h)2 +(3 – k)2 ie. h2 + k2 – 4h + 2k +5 = 4 + h2 –4h + 9 –6k + k2 or, 8k = 8 k = 1 again, Taking CP2 = CR2 ie. (2 – h)2 +(1 + k)2 = (4 – h)2 +(1 – k)2 ie. h2 + k2 – 4h + 2k +5 = h2 +k2 – 8h – 2k + 17 put k = 1, 4h + 4.1 = 12 or, 4h = 8 h = 2 centre (h,k) = (2, 1) Radius (r) = (x – h)2 +(y – k)2 ( (x,y)=(2,–1),(h,k)=(2,1) = (2 – 2)2 +(–1 – 1)2 = 2 units. Equation of the required circle is, (from (i)) (x – 2)2 +(y – 1)2 = 4
238 Vedanta Optional Mathematics Teacher's Guide ~ 10 or, x2 + y2 – 4x – 2y +5 = 4 x2 + y2 – 4x – 2y +1 = 0 is the required equation of the circle. Note: Alternatively above question can be solved by taking equation x2 + y2 +2gx + 2fy + c =0 putting the points and solving three obtained equations for g,f and c. 10. Find the equation of the circle passing through the points (5,7), (6,6) and (2,–2). Solution Let the required equation of the circle be x2 + y2 +2gx + 2fy + c = 0 ...........(i) The circle passes through the points (5,7), (6,6) and (2,–2), we get 52 + 72 +2. g. 5 + 2 .f .7 + c = 0 or, 10g + 14f + 74 + c = 0 .................... (ii) Similarly, 12g + 12f +72 +c =0 .............(iii) and 4g – 4f + 8 +c = 0 .............(iv) Subtracting (iii) from (ii), we get, –2g + 2f + 2 = 0 or, g – f = 1 ...........(v) Again subtracting (iv) from (iii) 8g + 16f + 64 =0 or, g + 2f + = –8 ................(vi) Solving equation (v) and (vi), we get, f = –3 and g =–2 centre = (–g, –f) = (2,3) Put the values of g and f in equation (ii), we get, 10×(–2) + 14 × (–3) + 74 + c =0 or, – 20 – 42 + 74 + c =0 c = –12 put the values of g, f and c in eqn(i), we get x2 + y2 – 4x – 6y – 12 = 0 which is the required equation. 11. Find the equation of the circle passing through the points (3,2) and (5,4) and centre lies on the line 3x – 2y = 1. Solution Let c (h,k) be the centre of the circle. It lies on the line 3x – 2y =1. ie. 3h – 2k = 1 ...........(1) Let A(3,2) and B (5,4) be the points on the circumference of the circle. Then
Vedanta Optional Mathematics Teacher's Guide ~ 10 239 CA = CB or, CA2 = CB2 (3 – h)2 +(2 – k)2 = (5 – h)2 +(4 – k)2 or, 9 – 6h + h2 + 4 – 4k + k2 = 25 – 10h +h2 + 16 – 8k + k2 or, 13 + 4h + 4k = 41 or, 4h + 4k = 28 h + k = 7 ............(ii) solving equation (i) and (ii), we get h = 3 and k = 4 centre = (h,k) =(3,4) Now, equation of the circle is (x – h)2 +(y – k)2 = r2 (x – 3)2 +(y – 4)2 = r2 where, r = (3 – 3)2 +(2 – 4)2 =2 or, x2 – 6x + 9 + y2 – 8y + 16 = 4 or, x2 – 6x + 9 + y2 – 8y + 16 = 4 x2 + y2 – 6x – 8y + 21 = 0 is the required equation of the circle. 12. Find the equation of the circle which touches the x –axis at (4, 0) and cuts off an intercepts of 6 units from the y – axis positively. Solution Let c(h,k) be the centre of the required circle. Let A(4,0) be the point on x –axis and DE = 6 units from y – axis. The point of (4,0) on the line 3x – 2y =1. ie. 3h – 2k = 1 ...........(1) Let A(3,2) and B (5,4) be the points on the circumference of the circle. Then Draw CB perpendicular on y – axis and CA OX . CA = k = radius = r h = BC = 4 untis = OA, BD = 3 Now, CD = K = BC2 +BD2 = 42 +32 = 25 = 5 centre = (h,k) =(4, 5) Now, equation of the required circle is given by (x – h)2 +(y – k)2 = k2 , r = k = 5 (x – 4)2 +(y – 5)2 = 52 or, x2 – 8x + 9 + y2 – 105 + 16 = 4 ⸫ x2 +y2 –8x–105+16=0 O D A(4,0) 4 C(h,k) 3 3 B E I x' x y y'
240 Vedanta Optional Mathematics Teacher's Guide ~ 10 13. Prove that the points A(2, –4), B(3, –1), C(3, –3) and D(0, 0) are concyclic. Solution Let P(h,k) be the centre of the circle and A(2, –4), B(3, –1), C(3, –3) and D(0, 0) be three points on the circumference of the circle. Then PA = PB = PC or, PA2 =PB2 = PC2 Taking PA2 = PB2 ie. (h – 2)2 +(k + 4)2 = (h – 3)2 +(k + 1)2 or, h2 –4h + 4 + k2 + 8k + 16 =h2 –6h + 9 + k2 +2k +1 or, 2h + 6k = –10 or, h + 3k = –5 ...............(i) Taking PB2 = PC2 ie. (h – 3)2 +(k + 1)2 = (h – 3)2 +(k + 3)2 or, k2 + 2k + 1 = k2 +6k +9 or, – 4k = 8 k = –2 Put the value of k in equation (i), we get h + 3(–2) = –5 or, h = 1 (h,k) =(1, –2) radius(r) =PA = (1 – 2)2 +(–2 + 4)2 = 1 +4 = 5 Now, equation of required circle is (x – h)2 +(y – k)2 = r2 (x – 1)2 +(y + 2)2 = 5 x2 + y2 – 2x + 4y = 0 ........(ii) put the point D(0,0) in equation (ii), we get, 0 = 0 (true) D(0,0) also lies on the circle. Hence the points A, B, C and D are concyclic. 14. Show that the two circles x2 + y2 = 36 and x2 + y2 – 12x – 16y + 84 = 0 touch externally. Solution Let the given circles be p1 and p2 as shown in the given figure. x2 + y2 = 36 ............(i) x2 + y2 – 12x – 16y + 84 = 0 ..........(ii) From equation (i), radius = r1 = 6, centre c1 = (0,0)
Vedanta Optional Mathematics Teacher's Guide ~ 10 241 Again, from circle (ii), radius (r2 ) = g2 + f2 – c , centre = c2 =(6,8) = (–6)2 + (–8)2 – 84 = 36 + 64 –84 = 16 = 4 units Distance between c1 and c2 = 62 + 82 = 100 = 10 units sum of the radii = r1 + r2 = 6 + 4 = 10 Since the distance between the centres of two circles is equal to the sum of radii of the circles. So given two circles touch externally. proved 15. Show that two circles touch internally. x2 + y2 = 81 and x2 + y2 – 6x – 8y + 9 = 0 . Solution Given equations of circles are x2 + y2 = 81 ............(i) x2 + y2 – 6x – 8y + 9 = 0 ..........(ii) From equation (i), radius = r1 = 9 imoyd, centre (h,k) = (0,0) Again, from equation (ii), radius = g2 + f2 – c = (–3)2 + (–4)2 = 9 + 16 = 25 = 5 units radius (r2 ) =5 centre = c2 =(3, 4) Difference of raddi = r1 – r2 = 9 – 5 = 4 Distance between the centres of the circles = 32 + 42 = 9 + 16 = 25 = 5 Since the distance between the centres of two circles is equal to the difference of the of radii of the circles, the two circles touch internally. proved Conic Section 16. Text book Q. N. 3(paper 184) Solution (a) (b) (c)Ellipse (d) Hyperbola P1 C1 P2 C2 r1 =9 r2 =5 C1 4 C2
242 Vedanta Optional Mathematics Teacher's Guide ~ 10 17. Find the equation of tangent to the circles x2 + y2 = 25 at (3, 4). Solution Let PT be the tangent to the circle x2 + y2 = 25............(i) at the point A(3, 4). Now, radius of circle = 5 centre = O(0,0) Slope OA (m2 )= y2 –y1 x2 –x1 = 4 – 0 3 – 0 =– 4 3 Let slope of tangent PT be m2 By plane geometry, we know that OA is perpendicular to PT m1 . m2 = –1 ie. mc = – 3 4 Now, equation of tangent PT is given by y – y1 =m(x – x1 ) where m= m2 =– 3 4 ie. y – 4 =– 3 4 (x – 3) or, 4y – 16 = –3x + 9 3x + 4y = 25 Note: Equation of tangent to the circles x2 + y2 =r2 is xx1 + yy1 = r2 Questions for practice 1. Find the equation of the circle with centre (2,3) and radius 5. 2. Find the centre and radius of the circle whose equation is x2 + y2 + 4x – 6y + 4=0 3. Find the equation of the circle whose centre is the point of intersection of x + 2y – 1 =0 and 2x – y –7 = 0 and it passes through (3,1). 4. Find the equation of the circle which touches positive axes of x and y and whose radius is 6 units. 5. Find the equation of the circle of the ends of a diameter are (3,2) and (7, –2). 6 (a). Find the equation of the circle passing through the points (1,0),(2,–2) and (3,1). (b). Find the equation of the circle through the points (1,2),(3,1) and (–3,–1). 7. Show that the points (3,3),(3,–3) and (–3,3) and (–3,–3) are concyclic. 8. Find the equation of the circle passing through the points (1, –5) and concentric with the circles, x2 + y2 – 4x – 8y – 81 = 0 (2,–2). 9. If y = x + 2 is the equation of a chord of the circle x2 + y2 + 2x = 0 . Find the equation of the circle of which this chord as an a diameter. Also find the length of the chord. 10. Find the equation of the tangent to the circle x2 + y2 =100 at (6,8). 11. Find the point of intersection of the line x + y = 3 and the circle x2 + y2 – 2x – 3 =0. Also find the length hord.(Ans: (1,2), (3,0), 2 2 ) 12. Show that the two circle touch externally. x2 + y2 – 4x – 6y – 3 = 0 and x2 + y2 – 22x – 4y + 109 = 0 A(3,4) O P T
Vedanta Optional Mathematics Teacher's Guide ~ 10 243 Multiple Angles Estimated Periods : 7 1. Objectives S,N. Level Objectives (i) Knowledge(K) To define multiple angles of angle A as 2A, 3A ect. To tell the formulae of trigonometric ratios of multiple angles. (ii) Understanding(U) To explain to derive the formulae of multiple angles by using compound angle formulae. (ii) Application(A) To solve problems of trigonometric identities of multiple angles. (iv) Higher Ability (HA) To interpreter sin2A, cos2A, tan2A geometrically. To solve trigonometric identities of difficult questions of multiple angles. 2. Teaching Materials Formula chart of compound angles and multiple angles. 3. Learning stategies – Review the formula of trigonometric ratios of compound angles studied in class 9. – Define multiple angles of A as 2A, 3A, ... ... ... – Show how to derive formulae of multiple angles of trigonometric ratios. eg. sin2A = 2 sinA . cos A, sin3A = 3 sinA – 4sin3 A – Discuss how to solve trigonometric identition wiht examples. 4 List of formulae 1. sin2A = 2 sinA . cosA = 2 tanA 1 + tan2 A = 2 cotA 1 + cot2 A 2. cos2A = cos2 A – sin2 A = 2 cos2 A – 1 = 1 – 2 sin2 A = 1 – tan2 A 1 + tan2 A = = cot2 A – 1 cot2 A + 1 3. tan2A = 2 tanA 1 – tan2 A 4. cot2A = cot2 A – 1 2 cotA 5.sin3A = 3 sinA – 4 sin3 A ⇒ 4 sin3 A = 3 sinA – sin3A 6. cos3A = 4 cos3 A – 3 cosA ⇒ 4 cos3 A = 3 cosA + cos3A 7. tan3A = 3 tanA – tan3 A 1 – 3 tan2 A 8. cot3A = cot3 A – 3 cotA 3 cot2 A – 1 nine Trigonometry UNIT
244 Vedanta Optional Mathematics Teacher's Guide ~ 10 Some solved problems 1. If cosθ = 1 2 p + 1 p , then show that: i) cos2θ = 1 2 p2 + 1 p2 ii) cos3θ = – 1 2 p3 + 1 p3 Solution i) Here, cos2θ = 2 cos2 θ – 1 = 2 1 4 p + 1 p 2 – 1 = 1 2 p2 + 2 . p . 1 p + 1 p2 – 1 = 1 2 p2 + 2 + 1 p2 – 1 = 1 2 p2 + 2 + 1 p2 – 2 = 1 2 p2 + 1 p2 ii) cos3θ = 4 cos3 θ – 3 cosθ = cosθ(4 cos2 θ – 3) = 1 2 p + 1 p 4 . 1 4 p + 1 p 2 – 3 = 1 2 p + 1 p p2 + 2 . p. 1 p + 1 p2 – 3 = 1 2 p + 1 p p2 + 2 + 1 p2 – 3 = 1 2 p + 1 p p2 – 1 + 1 p2 = 1 2 p + 1 p p2 – p . 1 p + 1 p2 = 1 2 p3 + 1 p3
Vedanta Optional Mathematics Teacher's Guide ~ 10 245 2. a) If cos θ = 6 5 2 , Show that cos2 θ = 11 25 . b. If tan θ = 5 12 , show that tan2 θ = 120 119 Solution i) Here, cos θ = 6 5 2 LHS = cos θ = 2cos 2 θ – 1 = 2 6 5 2 2 – 1 = 2 36 50 – 1 = 36 25 – 1 = 36 – 25 25 = 11 25 = RHS proved ii) Here, tan θ = 5 12 LHS = tan2 θ = 2 tan θ 1 – tan 2 θ = 2 . 5 12 1 – 25 144 = 56 × 144 144 – 25 = 120 119 = RHS p roved 3. Prove that following. a. cos2 θ 1 + sin2 θ = 1 – tan θ 1 + tan θ Solution LHS = cos2 θ 1 + sin2 θ = cos 2 θ – sin 2 θ sin 2 θ + cos 2 θ + 2 sinθ cos θ
246 Vedanta Optional Mathematics Teacher's Guide ~ 10 = (cosθ + sinθ) (cosθ – sinθ) (cosθ + sinθ) 2 = cosθ – sinθ cosθ + sinθ (dividing numerator and denominator by cosθ) = 1 – tanθ 1 + tanθ = RHS proved Alternate Method LHS = cos2θ 1 + sin2θ = 1 – tan2 θ 1 + tan2 θ 1 + 2 tanθ 1 + tan2 θ = 1 – tanθ 1 + tanθ = RHS proved b. cosθ cosθ – sinθ – cosθ cosθ + sinθ = tan2θ Solution LHS = cosθ cosθ – sinθ – cosθ cosθ + sinθ = cosθ(cosθ + sinθ) – cosθ(cosθ – sinθ) (cosθ – sinθ) (cosθ + sinθ) = cos2 θ + sinθ . cosθ – cos2 θ + sinθ . cosθ cos2 θ – sin2 θ = 2 sinθ . cosθ cos2θ = sin2θ cos2θ = tan2θ = RHS proved c. 1 + sin2A cos2A = sinA + cosA cosA – sinA Solution LHS = 1 + sin2A cos2A = sin2 A + cos2 A + 2 sinA . cosA cos2 A – sin2 A = (cosA + sinA)2 (cosA + sinA) (cosA – sinA) = sinA + cosA cosA – sinA = RHS proved
Vedanta Optional Mathematics Teacher's Guide ~ 10 247 d. sin5 θ sin θ – cos5 θ cos θ = 4 cos2θ Solution LHS = sin5 θ sin θ – cos5 θ cos θ = cos θ . sin5 θ – sin θ .cos5 θ sin θ . cos θ = sin(5 θ – θ ) sin θ . sin θ = sin4 θ sinθ . cos θ = sin(2θ) sinθ . cos θ = 2sin2θ . cos2 θ sinθ . cos θ = 4 sinθ . cosθ . cos2 θ sinθ . cos θ = 4 cos2θ = RHS proved e. cos2 θ 1 + sin2 θ = tan π4 – θ Solution LHS = cos2 θ 1 + sin2 θ = 1 – tan 2 θ 1 + tan 2 θ 1 + 2 tan θ 1 + tan 2 θ = 1 – tan 2 θ 1 + tan 2 θ × 1 + tan 2 θ 1 + tan 2 θ + 2 tan θ = (1 – tan θ) (1 + tan θ ) (1 + tan θ ) 2 = 1 – tan θ 1 + tan θ = tan π4 – tan θ 1 + tan π4 . tanθ = tan π4 – θ = RHS proved 4. Prove that following.
248 Vedanta Optional Mathematics Teacher's Guide ~ 10 a. sin2θ – cosθ 1 – sinθ – cos2θ = cotθ Solution LHS = sin2θ – cosθ 1 – sinθ – cos2θ = 2 sinθ . cosθ – cosθ 1 – sinθ – 1 + 2 sin2 θ = cosθ(2 sinθ – 1) sinθ(2 sinθ – 1) = cotθ = RHS proved b. 1 + sin2θ – cos2θ 1 + sin2θ + cos2θ = tanθ Solution LHS = 1 + sin2θ – cos2θ 1 + sin2θ + cos2θ = (1 – cos2θ) + sin2θ (1 + cos2θ) + sin2θ = 2 sin2 θ + 2 sinθ . cosθ 2 cos2 θ + 2 sinθ . cosθ = 2 sinθ(sinθ + cosθ) 2 cosθ(sinθ + cosθ) = sinθ cosθ = tanθ = RHS proved c. sinθ + cosθ cosθ – sinθ + cosθ – sinθ cosθ + sinθ = 2 sec2θ Solution LHS = sinθ + cosθ cosθ – sinθ + cosθ – sinθ cosθ + sinθ = (sinθ + cosθ) 2 + (cosθ – sinθ) 2 (cosθ – sinθ) (cosθ + sinθ) = sin2 θ + 2 sinθ . cosθ + cos2 θ + cos2 θ – 2 sinθ . cosθ + sin2 θ cos2 θ – sin2 θ = 2(sin2 θ + cos2 θ) cos2 θ – sin2 θ = 2 cos2θ = 2 sec2θ = RHS proved
Vedanta Optional Mathematics Teacher's Guide ~ 10 249 d. (1 + sin2 θ + cos2 θ ) 2 = 4 cos 2 θ(1 + sin2 θ ) Solution LHS = (1 + sin2 θ + cos2 θ ) 2 = (1 + 2 sin θ . cos θ + 2 cos 2 θ – 1) 2 = [2 cos θ(sin θ + cos θ)] 2 = 4 cos 2 θ(sin θ + cos θ ) 2 = 4 cos 2 θ(sin 2 θ + 2 sin θ . cos θ + cos 2 θ ) = 4 cos 2 θ(1 + sin2 θ) = RHS proved e. 1 tan2 θ – tan θ – 1 cot2 θ – cot θ = cot θ Solution LHS = 1 tan2 θ – tan θ – 1 cot2 θ – cot θ = 1 tan2 θ – tan θ – 1 1 tan2 θ – 1 tan θ = 1 tan2 θ – tan θ – tan2θ . tan θ tanθ – tan2 θ = 1 + tan2 θ . tan θ tan2 θ – tan θ = 1 + 2 tan θ 1 – tan 2 θ tan θ 2 tan θ 1 – tan 2 θ – tan θ = 1 – tan 2 θ + 2tan 2 θ 2 tanθ – tanθ + 1 – tan 3 θ = 1 + tan 2 θ tanθ + tan 3 θ = 1 + tan 2 θ tan θ(1 + tan 2 θ ) = 1 tan θ = cot θ = RHS proved 5. Prove that: a. 1 – sin2 θ 1 + sin2 θ = cotθ – 1 cotθ + 1 2
250 Vedanta Optional Mathematics Teacher's Guide ~ 10 Solution LHS = 1 – sin2θ 1 + sin2θ = 1 + 2 tanθ 1 + tan2 θ 1 + 2 tanθ 1 + tan2 θ = 1 + tan2 θ – 2 tanθ 1 + tan2 θ + 2 tanθ = 1 + tanθ 1 – tanθ 2 = 1 – 1 cotθ 1 + 1 cotθ 2 = cotθ – 1 cotθ + 1 2 = RHS proved b. 1 + tan2 π 4 – θ 1 – tan2 π 4 – θ = cosec2θ Solution LHS = 1 + tan2 π 4 – θ 1 – tan2 π 4 – θ = 1 1 – tan2 π 4 – θ 1 + tan2 π 4 – θ = 1 cos π 4 – θ = 1 cos π 2 – 2θ = 1 sin2θ = cosec2θ = RHS proved